# FP4 Factorising Determinants

Thread starter 7 years ago
#1
Hi all,

I am stuck on this question about factorising determinants. I can't find a common factor in any of the lines or columns. I've tried adding and subtracting rows/columns, but to no avail. The a3, b3 and c3 particularly confuse me here.
1
7 years ago
#2
(Original post by timgb)
hi all,

i am stuck on this question about factorising determinants. I can't find a common factor in any of the lines or columns. I've tried adding and subtracting rows/columns, but to no avail. The a3, b3 and c3 particularly confuse me here.

a3 - b3 = (a - b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 - ab + b2)
0
Thread starter 7 years ago
#3
(Original post by TeeEm)

a3 - b3 = (a - b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 - ab + b2)
Ah, that factorises nicely. Thank you for the help!

(a+b+c)(a-b)(b-c)(c-a)
0
7 years ago
#4
(Original post by TimGB)
Ah, that factorises nicely. Thank you for the help!
my pleasure
0
Thread starter 7 years ago
#5
(Original post by TeeEm)
my pleasure
I would give you a thumbs up, but it seems TSR won't let me. Probably because I gave you one for helping me 3 or so months ago.
0
7 years ago
#6
(Original post by TimGB)
I would give you a thumbs up, but it seems TSR won't let me. Probably because I gave you one for helping me 3 or so months ago.
You should use your thumbs up more regularly.
I have a rep list and I rep every helper in this room on a cyclic basis (like your determinant).
Not to worry about it.
0
7 years ago
#7
(Original post by TimGB)
Hi all,

I am stuck on this question about factorising determinants. I can't find a common factor in any of the lines or columns.
As an alternative method:

If you consider the determinant as a polynomial in "a", then setting a equal to b gives two identical rows and the determinant is zero. By the factor theorem, (a-b) is a factor....
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